Question

有这样的字符串：

let str = "In 1273, however, they lost their son in an accident;[2] the young Theobald was dropped by his nurse over the castle battlements.[3]"

我正在寻找一种解决方案，删除方括号的所有外观以及它之间的任何东西。

我正在尝试使用String的方法：replacingOccurrences(of:with:)，但它需要删除所需的确切子字符串，所以它对我不起作用。

Answer 1

您可以使用：

let updated = str.replacingOccurrences(of: "\\[[^\\]]+\\]", with: "", options: .regularExpression)

正则表达式（没有Swift字符串中所需的转义符：

\[[^\]+]\]

\[和\]查找字符[和]。他们有一个反斜杠来删除正则表达式中这些字符的正常特殊含义。

[^]]表示匹配除]字符以外的任何字符。 +表示匹配1或更多。

Answer 2

您可以创建一个while循环来获取第一个字符串范围的lowerBound和第二个字符串范围的upperBound，并从中创建一个范围。接下来只删除字符串的子范围，并为搜索设置新的startIndex。

var str = "In 1273, however, they lost their son in an accident;[2] the young Theobald was dropped by his nurse over the castle battlements.[3]"

var start = str.startIndex

while let from = str.range(of: "[", range: start..<str.endIndex)?.lowerBound,
    let to = str.range(of: "]", range: from..<str.endIndex)?.upperBound,
    from != to {
        str.removeSubrange(from..<to)
        start = from
}

print(str)   // "In 1273, however, they lost their son in an accident; the young Theobald was dropped by his nurse over the castle battlements."

知道Swift中第一个和最后一个字符的字符串中的子串

2 个答案: